User:Matt Kwan/Newton's Method

to prove: if $$f(a)<0, f(b)>0, f'(x)>0, f''(x)>0\!$$, then Newton's method $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ converges; $$\lim_{n\rightarrow \infty}x_n=r$$ such that f(r)=0.

$$ \begin{align} g_n(x)&=f'(x_n)(x-x_n)+f(x_n)\\ g_n(x_n+h)&=hf'(x_n)+f(x_n)\\ \frac{f(x_n+h)-f(x_n)}h&=f'(c), c>x_n\\ f(x_n+h)&=hf'(c)+f(x_n)\\ &>g_n(x_n+h) \end{align} $$

Now define $$h_1$$ and $$h_2$$ such that $$g_n(x_n+h_1)=0, f(x_n+h_2)=0\!$$. If $$h_1\le h_2$$, then $$g_n(x_n+h_1)=f(x_n+h_1)+k, k\ge 0$$, so $$g_n(x_n+h_1)\ge f(x_n+h_1)$$, a contradiction. So $$h_1>h_2, x_n+h>r\!$$.

$$g_n(x_n+h)=0$$ implies $$x_n+h=x_n-\frac{f(x_n)}{f'(x_n)}$$ so $$f\left( x_n-\frac{f(x_n)}{f'(x_n)}\right)=f(x_{n+1})>f(r)>0$$.

$$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ so $$x_{n+1}0 so $$f(x_{n+1})f(x2)>\cdots >0$$ so $$x_1>x2>\cdots >r$$ so by the monotone sequence theorem, $$x_n$$ converges. So $$\lim_{n\rightarrow \infty}\left(x_n-x_{n+1}\right)=0$$. But $$f(x_n)=f'(x_n)(x_n-x_{n-1})\!$$ and $$f'(x_n)\le \max \left \{ f'(x):r\le x\le x_1 \right \}$$ so $$\lim_{n\rightarrow \infty}f(x_n)=0$$ so $$x_n \rightarrow r$$ as required.